Help to understand a step in the fix point theorem

Question

Let $G$ be a $p$-group and $X$ be a finite set on which G operates. We define $X_G:=\{x \in X: g \circ x=x \; \forall g \in G\}$ set the of all fixed points. For the proof of $|X|\equiv |X_G|$ mod $p$ I don't understand that step:

$x \in X_G \; \Rightarrow (G:\operatorname{Stab}(x))=1$

Why is this true? Thanks for your help

Answer 1

Let $G$ be a group. Then

$$X_G=\{x\in X\ |\ gx=x\text{ for any }g\in G\}$$ $$Stab(x)=\{g\in G\ |\ gx=x\}$$

We will show that $x\in X_G$ if and only if $Stab(x)=G$.

"$\Rightarrow$" We need to show that any element of $G$ is in fact in $Stab(x)$. So let $g\in G$. Since $x\in X_G$ then $gx=x$ and thus $g\in Stab(x)$. $\Box$

"$\Leftarrow$" Let $x\in X$. We need to show that $x\in X_G$. So let $g\in G$. Since $Stab(x)=G$ then $gx=x$. By the arbitrary choice of $g$ we get that $x\in X_G$. $\Box$

Answer 2

Definition of stabilizer:

$$\operatorname{Stab}(x):=\{g\in G\mid g\cdot x=x\} \tag 1$$

If $x \in X_G$, then $g\cdot x=x, \forall g\in G$; this latter condition can be rewritten as:

$$g \in G\space\Rightarrow\space g\cdot x=x\space\Rightarrow\space g \in \operatorname{Stab}(x) \tag 2$$

whence: $x\in X_G \Rightarrow G\subseteq \operatorname{Stab}(x)\Rightarrow G=\operatorname{Stab}(x)$.